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mestny [16]
3 years ago
7

What is the measure of c? a = 15 j = 24 k = 32

Mathematics
1 answer:
inessss [21]3 years ago
4 0

Answer:

The answer to your question is c = 20

Step-by-step explanation:

Data

a = 15

j = 24

k = 32

c = ?

Process

Use proportions to solve this problem. Compare the small triangle and the large triangle.

1.- Proportion of the small triangle

             c/15

2.- Proportion of the large triangle

            k/j

3.- Equal but terms and solve for c

            c/15 = k/j

-Substitution

            c / 15 = 32/24

            c = 32(15)/24

-Result

            c = 20

       

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Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with six rectangles and use the the
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Answer:

Summation notation:

\frac{1}{2}\sum_{k=1}^6f((.5k))

or after using your function part:

\frac{1}{2}\sum_{k=1}^6(-(.5k)^2+2(.5k)+4)

After evaluating you get 11.125 square units.

Step-by-step explanation:

The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.

The distance between x=0 and x=3 is (3-0)=3.

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We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).

The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).

The next will be at x=2.5-.5=2, and the third rectangle will have  a height of f(2).

The fourth rectangle will have a height of f(2-.5)=f(1.5).

The fifth one will have a height of f(1.5-.5)=f(1).

The last one because it is the sixth one will have a height of f(1-.5)=f(.5).

So to find the area of a rectangle you do base*time.

So we just need to evaluate:

\frac{1}{2}f(3)+\frac{1}{2}f(2.5)+\frac{1}{2}f(2)+\frac{1}{2}f(1.5)+\frac{1}{2}f(1)+\frac{1}{2}f(.5)

or by factoring out the 1/2 part:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

To find f(3) replace x in -x^2+2x+4 with 3:

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To find f(2.5) replace x in -x^2+2x+4 with 2.5:

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To find f(2) replace x in -x^2+2x+4 with 2:

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To find (1.5) replace x in -x^2+2x+4 with 1.5:

-1.5^2+2(1.5)+4

-2.25+3+4

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To find f(1) replace x in -x^2+2x+4 with 1:

-1^2+2(1)+4

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To find f(.5) replace x in -x^2+2x+4 with .5:

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\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

is it

\frac{1}{2}\sum_{k=1}^{6}(f(.5+.5(k-1)))

or after simplifying a bit:

\frac{1}{2}\sum_{k=1}^6 f((.5+.5k-.5))

\frac{1}{2}\sum_{k=1}^6f((.5k))

If you are wondering how I obtain the .5+.5(k-1):

I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)

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